Determining $x + y + z$

Question

$x$, $y$ and $z$ are different prime numbers;

$$x(z-y) = 18$$

$$y(z-x) = 40$$

Determine $x+y+z$.

I conceive of a way to solve this problem, which is to analyze the cases. For instance, Let's recall $x = 2$, then $z-y$ should be $9$. It will also give us $y = 5$. However, it doesn't satisfy the condition due to $z = 14$, which is not a prime number. How would you solve this question?

Regards!

Answer 1

Hint: $40 = 2^3 \times 5$. So $y$ must be $2$ or $5$. Similarly $x=2,3$.

Answer 2

Given that the are different integers and prime the $x(z-y) = 18$ means $x = 2$ or $x=3$. $z > y$ and $z -y = 9$ or $z-y= 6$.

$y(z-x) =40$ mean $y = 2$ or $5$. $z> x$ and $z-x = 20$ or $z-x =8$.

There are only so many cases to check.

1: $x =2; y=5$; So $z-5=18;z-2 = 8$ so $z = 23=10$. This is impossible.

2: $x=3; y = 2$; So $z-2 = 6; z-3= 20$ so $z = 8 = 23$. This is impossible.

3: $x=3; y = 5$; So $x-6 =6; z-3=8$ so $z = 11$. This is possible.

As those are the only three options $x = 3; y = 5; z = 11$.

Answer 3

I found another method:

Simply distribute the x and y in each expression.You will find that if you add the two resulting equations together, you will get $xz - yz = -22$, or $(x-y)z = -22.$ Now you know that y > x, and z must be 2 or 11. It should be relatively easy to try out just two cases instead of randomly plugging in numbers.